Proof that closure of a space equals another space.

[Paalfaal]

Homework Statement

Define:

- c₀ = {(x_n)_n $\in$ $\ell$_$\infty$ : lim_{n → $\infty$} x_n = 0}

- l₀ = {(x_n)_n $\in$ $\ell$_$\infty$ : $\exists$ N $\in$ the natural numbers, (x_n)_n = 0, n $\geq$ N}Problem: Prove that $\overline{\ell}$₀= c₀ in $\ell$$\infty$

Homework Equations

The Attempt at a Solution

I want to find the solution using the limit-definition of closure.

Considering an element

x = (x_n) $\in$ c₀

and a sequence

y^j = (x^j_n) $\in$ $\ell$₀,

such that x_n^j = x_n for n < j, x_n^j = 0, otherwise.

Using the metric induced by the supremum norm; || x_n - x_n^j ||_∞ $\rightarrow$ 0 as j tends to infinity. We can du this for all elements in c₀, and hence c₀ $\subseteq$ $\overline{\ell}$₀.

My problem is to show the other direction, that is

$\overline{\ell}$₀ $\subseteq$ c₀

I need to show that elements in $\overline{\ell}$₀ has a vanishing limit. I don't know how to do this using the supremum norm. In fact, it seems impossible to me..

Can I get any help?

 

[micromass]

Clearly, we know that $\ell_0\subseteq c_0$. Try to prove now that $c_0$ is closed. This would imply what you try to prove.

 

[Paalfaal]

Hmmm.. When is it true that $A \subseteq B \Rightarrow$ $\bar{A}$ $\subseteq$ $\bar{B}$ ?


I ask because I've been working on a similar problem: Prove that $\bar{\ell}$₀ = $\ell$₂ in $\ell$₂.

With a similar approach of the attempt above I got $\bar{\ell}$₀ $\supseteq$ $\ell$₂.

Since $\ell$₀ $\subseteq$ $\ell$₂ (for obvious reasons),
$\bar{\ell}$₀ $\subseteq$$\ell$₂ iff $\ell$₂ = $\bar{\ell}$₂ (correct me if I'm wrong here).

In that case, $\bar{\ell}$₀ = $\ell$₂ in $\ell$₂.


(I guess it might is trivial that $\ell$₂ = $\bar{\ell}$₂ in $\ell$₂, but Is there an easy way to prove this?)

Any comments?

 

[micromass]

Hmmm.. When is it true that $A \subseteq B \Rightarrow$ $\bar{A}$ $\subseteq$ $\bar{B}$ ?

This is always true. Try to prove it.

(I guess it might is trivial that $\ell$₂ = $\bar{\ell}$₂ in $\ell$₂, but Is there an easy way to prove this?)

If X is a metric space, then $\overline{X}=X$ (closure in X). Try to prove this.

 

[Paalfaal]

Try to prove it.

I certiantly will!


Thank you for your help! Much appreciated